Dome za balance when ch at stow (usng td tensions)

• bring the telescope into focus using tietrk  then turn off the tiedowns (so they don't move during the test).
• move the carriage house to stow position (8.835 degrees).
• swing the azimuth from 90 <--> 450 degrees at +/- .38 degs/sec. Do this with the dome at 7,7.5,8,8.5 degrees.
• For each azimuth swing fit y=c0 + c1*cos(az) + c2*sin(az).
• Compute the amplitude of the 1a term: sqrt(c1^2 + c2^2)
• evaluate the fit at the azimuth of each tiedown.

• Top: the 1 az amplitude for the azwings at za=7,7.5,8,8.5.
• black:td12, red:td4,green:td8
• It goes to 0 close to za=8
• Bottom: evaluated each 1az fit at the tiedown azimuth.
• the tension goes from positive (dome side to light) to negative (dome side to heavy) at the balance point.
• A linear fit to all the points vs za gives a balance of za=7.968 degress (optic axis).

Summary:

• With the ch at stow, the azimuth arm  balances with dome za=7.968 degrees.
• Saying that the azimuth is balanced does not mean that all 3 tiedowns have the same tensions, just that the tensions don't change as you do an az swing.
• Tiedowns will have an unequal tension since the weight distribution on the triangle is not uniform (there is more weight in corner 8).
  
• Using the expected weights, distances for the various things gave a balance of 7.5 degrees:
• dome=215kips
• ch=35kips
• chCounterWeight=45 kips
• chCounterWeightdistance:145 feet
• Radius for dome,ch : 430 feet.
• dome enter of mass 1.9 degrees downhill from the encoder value.
• What is missing is the residual unbalance from the  az arm stairway, vertex shelter, etc.. see computing unbalanced loads: